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Is there a formula to calculate the standard deviation of a group or individual statistics

Example in a sporting fantasy league I can calculate the summary statistics of any one player.

Example

one player has a Mean 75, Standard deviation 26.

But what if I selected a group of 18 players with the same summary stats, can I estimate the summary statistics of this group?

The Mean I believe can just be multiplied 1350 (75*18),

but the standard deviation isn't as simple as using a multiplication?

To add to this- if the group of 18 players each had unique summary stats, can the groups summary stats also be explained/calculated? example

Player 1 - mean 75 sd 26 Player 2 - mean 100 sd 28 Player 3 - mean 105 sd 25 . . Player 18 - mean 72 sd 23

Any help would be great.

Thanks heaps

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Well, it's complicated, because it depends on the distribution of the underlying statistic for each player. Let's assume we are talking about normal distributions. In that case (if the sample is large enough, you can ignore this constraint, and despite most data not being normally distributed, it's often somewhat close)

  • Means add up, so if player $A$ has mean $\mu_a$ and player $B$ has mean $\mu_b$, the mean of their sum is $\mu_a+\mu_b$. You can therefore multiply your common mean by 18 (because that's just adding up the 18 means)
  • Standard deviations don't add up, but their squares (variances) do! So, if standard deviation for player A is $\sigma_a$ and standard deviation for player B is $\sigma_b$, the standard deviation for their sum is $\sqrt{\sigma_a^2 + \sigma_b^2}$. In your case, square the standard deviations to get variances, add them up (multiply by 18) and then take the square root to have a standard deviation rather than a variance again
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  • $\begingroup$ Thanks David that makes perfect sense. $\endgroup$
    – Thomas Mason
    Jun 17, 2019 at 11:31
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    $\begingroup$ I am assumimg the data is normally distributed. Just to confirm - can I use this method for the calculation of the group of standard dev, even if the means are different? i.e. the means and variances of any group can be added as long as the distribution of the underlying stat is normally distributted. $\endgroup$
    – Thomas Mason
    Jun 17, 2019 at 11:35
  • $\begingroup$ Very good question! Yes, you can! If $X$ follows a $N(\mu_x, \sigma_x^2)$, Y follows $N(\mu_y, \sigma_y^2)$ and $X$ is independent from $Y$, then $X+Y$ follows a $N(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$ (see en.wikipedia.org/wiki/…) so these properties still apply even if means are different $\endgroup$
    – David
    Jun 17, 2019 at 11:39

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